Kenneth Kuttler

18 CHAPTER 2. THE REAL AND COMPLEX NUMBERS

10. Solve 3x+9x2+2x+1 ≥ 1.

11. Solve x2+2x+13x+7 < 1.

12. Solve |x+1|= |2x−3| .

13. Solve |3x+1| < 8. Give your answerin terms of intervals on the real line.

14. Sketch on the number line the solu-tion to the inequality |x−3|> 2.

15. Sketch on the number line the solu-tion to the inequality |x−3|< 2.

16. Show |x|=√

x2.

17. Solve |x+2|< |3x−3| .

18. Tell when equality holds in the trian-gle inequality.

19. Solve |x+2| ≤ 8+ |2x−4| .

20. Solve (x+1)(2x−2)x≥ 0.

21. Solve x+32x+1 > 1.

22. Solve x+23x+1 > 2.

23. Describe the set of numbers, a suchthat there is no solution to |x+1| =4−|x+a| .

24. Suppose 0 < a < b. Show a−1 > b−1.

25. Show that if |x−6|< 1, then |x|< 7.

26. Suppose |x−8| < 2. How large can|x−5| be?

27. Obtain a number, δ > 0, such that if|x−1|< δ , then

∣∣x2−1∣∣< 1/10.

28. Obtain a number, δ > 0, such that if|x−4|< δ , then |

√x−2|< 1/10.

29. Suppose ε > 0 is a given positivenumber. Obtain a number, δ >0, such that if |x−1| < δ , then|√

x−1| < ε . Hint: This δ will de-pend in some way on ε. You need totell how.

2.6 The Binomial TheoremConsider the following problem: You have the integers Sn = {1,2, · · · ,n} and k is an integerno larger than n. How many ways are there to fill k slots with these integers starting fromleft to right if whenever an integer from Sn has been used, it cannot be re used in anysucceeding slot?

k of these slots︷︸︸︷, , , , · · · ,

This number is known as permutations of n things taken k at a time and is denoted byP(n,k). It is easy to figure it out. There are n choices for the first slot. For each choicefor the fist slot, there remain n− 1 choices for the second slot. Thus there are n(n−1)ways to fill the first two slots. Now there remain n−2 ways to fill the third. Thus there aren(n−1)(n−2) ways to fill the first three slots. Continuing this way, you see there are

P(n,k) = n(n−1)(n−2) · · ·(n− k+1)

ways to do this.Now define for k a positive integer, k! ≡ k (k−1)(k−2) · · ·1, 0! ≡ 1. This is called

k factorial. Thus P(k,k) = k! and you should verify that P(n,k) = n!(n−k)! . Now consider

the number of ways of selecting a set of k different numbers from Sn. For each set of knumbers there are P(k,k) = k! ways of listing these numbers in order. Therefore, denotingby(n

)the number of ways of selecting a set of k numbers from Sn, it must be the case that(n

)k! = P(n,k) = n!

(n−k)! . Therefore,(n

)= n!

k!(n−k)! . How many ways are there to select no

18 CHAPTER 2. THE REAL AND COMPLEX NUMBERS3x+9 x4210. Solve Deal >1. 22. Solve 355 > 2.11. Solve edetl <l. 23. Describe the set of numbers, a suchthat there is no solution to |x+1| =12. Solve |x+ 1] = |2x—3]. 4—|x+al.13. Solve |3x+ 1] < 8. Give your answer 94. Suppose 0 <a <b. Showa! > b~!.in terms of intervals on the real line.; 25. Show that if |x—6| < 1, then |x| <7.14. Sketch on the number line the solu-tion to the inequality |x—3]| > 2. 26. Suppose |x—8| < 2. How large can15. Sketch on the number line the solu- [v5] be?tion to the inequality |x — 3| < 2. 27. Obtain a number, 6 > 0, such that if16. Show |x| = Vx2. |x—1| <6, then |x? — 1] < 1/10.17. Solve |x+2| < |3x—3]. 28. Obtain a number, 6 > 0, such that if|x —4| < 6, then |,/x—2| < 1/10.18. Tell when equality holds in the trian-gle inequality. 29. Suppose € > 0 is a given positivenumber. Obtain a number, 6 >19. Solve |x+2| <8 + |2x—4]. 0, such that if |x—1| < 6, then|,/x—1| < €. Hint: This 6 will de-pend in some way on €. You need to21. Solve a3 >1. tell how.20. Solve (x+1)(2x—2)x> 0.2.6 The Binomial TheoremConsider the following problem: You have the integers S,, = {1,2,--- ,n} and k is an integerno larger than n. How many ways are there to fill & slots with these integers starting fromleft to right if whenever an integer from S, has been used, it cannot be re used in anysucceeding slot?k of these slotsThis number is known as permutations of n things taken k at a time and is denoted byP(n,k). It is easy to figure it out. There are n choices for the first slot. For each choicefor the fist slot, there remain n — 1 choices for the second slot. Thus there are n(n— 1)ways to fill the first two slots. Now there remain n — 2 ways to fill the third. Thus there aren(n—1)(n—2) ways to fill the first three slots. Continuing this way, you see there areP(n,k) =n(n—1)(n—2)---(n—k+1)ways to do this.Now define for k a positive integer, k! = k(k—1)(k—2)---1, 0! = 1. This is calledk factorial. Thus P(k,k) =k! and you should verify that P (n,k) = can Now considerthe number of ways of selecting a set of k different numbers from S,. For each set of knumbers there are P (k,k) = k! ways of listing these numbers in order. Therefore, denotingby (7h) the number of ways of selecting a set of k numbers from S,,, it must be the case that(kK! = P(n,k) = cae Therefore, (7) = Cae How many ways are there to select no